MathsICSE Class 10 · Section and Mid-point Formula

Centroid of a Triangle: Formula, Medians and Solved Questions (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

The centroid G of a triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is G((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3), the average of the three vertices. It is where the three medians meet, and it divides each median in the ratio 2 : 1 from the vertex.

The three medians meet at G, which sits two-thirds of the way from each vertex, so AG : GD = 2 : 1 and G is the average of the vertices.

Illustration drawn for QuestionBuddy.

Understand Centroid of a Triangle

The CISCE syllabus includes the coordinates of the centroid of a triangle within the section and mid-point formula chapter.

What the centroid is. A median joins a vertex to the mid-point of the opposite side. The three medians of a triangle meet at one point, the centroid $G$ . It is the balance point of a flat triangular plate of even thickness.

The formula.

$G = (\frac{x _{1} + x _{2} + x _{3}}{3}, \frac{y _{1} + y _{2} + y _{3}}{3}) .$

Add the three x-coordinates and divide by 3; do the same for $y$ . Write the answer as a point, $(x, y)$ . In 2024 Q7(iii) the examiners reported candidates who wrote the centroid as $(\frac{0}{3}, \frac{0}{3})$ or as ' $x = 0$ , $y = 0$ ' instead of simplifying to $(0, 0)$ .

Where the formula comes from. $G$ lies on median $A D$ with $A G : G D = 2 : 1$ . With $D = (\frac{x _{2} + x _{3}}{2}, \frac{y _{2} + y _{3}}{2})$ , the section formula gives

$x_{G} = \frac{2 \cdot \frac{x _{2} + x _{3}}{2} + 1 \cdot x _{1}}{3} = \frac{x _{1} + x _{2} + x _{3}}{3} .$

So two routes give the same point: average the vertices, or go two-thirds of the way down any median.

Working backwards. If $G$ and two vertices are known, the third vertex is $C = (3 x_{G} - x_{1} - x_{2}, 3 y_{G} - y_{1} - y_{2})$ . If the vertices contain letters, equate each average to the matching coordinate of $G$ .

Useful facts.

$A G = \frac{2}{3} A D$ and $G D = \frac{1}{3} A D$ , so $A G = 2 G D$ .
The triangle formed by joining the mid-points of the sides has the same centroid as the original triangle.
The centroid always lies inside the triangle.

Centroid of a Triangle: formulas to remember

Centroid: $G = (\frac{x _{1} + x _{2} + x _{3}}{3}, \frac{y _{1} + y _{2} + y _{3}}{3})$; Average of the vertices.
Centroid on a median: $A G : G D = 2 : 1$; $D$ is the mid-point of $B C$ .
Third vertex: $C = (3 x_{G} - x_{1} - x_{2}, 3 y_{G} - y_{1} - y_{2})$; When $G$ , $A$ and $B$ are known.
Centroid from vertex and mid-point: $G = (\frac{2 x _{D} + x _{A}}{3}, \frac{2 y _{D} + y _{A}}{3})$; Section formula with $2 : 1$ .

How to solve Centroid of a Triangle questions

Add and divide by 3, or use the 2 : 1 split of a median.

1List the three vertices with their signs.
2Add the x-coordinates and divide by 3; add the y-coordinates and divide by 3.
3Write the answer as a single point and simplify.
4If only a vertex and the mid-point of the opposite side are known, use the section formula with ratio $2 : 1$ from the vertex.
5For a missing vertex, use $3 G$ minus the two known vertices.
6Check: the centroid should lie inside the triangle.

Centroid of a Triangle in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Centroid as a fair meeting point

Suppose three friends live at $(1, 7)$ , $(9, 3)$ and $(2, - 1)$ on a town grid (in km) and decide to meet at the centroid of the triangle formed by their homes.

Where do they meet?

Try it first, then show the solutionSolution

$x = \frac{1 + 9 + 2}{3} = 4$ .
$y = \frac{7 + 3 + ( - 1 )}{3} = 3$ .

Answer: At $(4, 3)$ .

Memory hook: Three homes, divide by three.

Centroid as a balance point

Imagine a triangular metal signboard of even thickness with corners at $(0, 0)$ , $(9, 0)$ and $(3, 6)$ (in dm), to be hung from a single hook so that it balances.

Where should the hook be attached?

Try it first, then show the solutionSolution

A uniform triangular plate balances at its centroid.
$G = (\frac{0 + 9 + 3}{3}, \frac{0 + 0 + 6}{3}) = (4, 2)$ .

Answer: At $(4, 2)$ .

Memory hook: Centroid = centre of balance.

Tricks to remember Centroid of a Triangle

Short memory aids. Say each one aloud once, then use it on a question the same day.

Mid-point ÷ 2, centroid ÷ 3
Two points, average over 2; three vertices, average over 3.
Use it when: Keeping the two formulas apart.
Two-thirds down the median
From a vertex, the centroid is $\frac{2}{3}$ of the way to the opposite mid-point.
Use it when: Questions that give a vertex and a mid-point.
Three G minus the rest
Missing vertex $= 3 G - (sum of the other two)$ .
Use it when: Finding a vertex from the centroid.

Centroid of a Triangle: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: A direct centroid · 1 mark · MCQBoard level

The centroid of the triangle with vertices $(1, 2)$ , $(3, - 4)$ and $(5, 8)$ is:
(a) $(9, 6)$
(b) $(3, 2)$
(c) $(4.5, 3)$
(d) $(3, 6)$

Try it first, then show the solutionSolution

$(\frac{1 + 3 + 5}{3}, \frac{2 - 4 + 8}{3}) = (3, 2)$ . Option (a) forgot to divide by 3.

Answer: (b) $(3, 2)$

Example 2: Centroid with negative coordinates · 2 marks · Short answerBoard level

Find the centroid of the triangle with vertices $A (- 2, 3)$ , $B (4, - 1)$ and $C (7, 7)$ .

Try it first, then show the solutionSolution

$x = \frac{- 2 + 4 + 7}{3} = 3$ .
$y = \frac{3 - 1 + 7}{3} = 3$ .

Answer: $G (3, 3)$

Example 3: The missing vertex · 3 marks · Short answerHarder board level

The centroid of triangle $A B C$ is $G (1, - 2)$ . If $A$ is $(3, 1)$ and $B$ is $(- 2, - 4)$ , find $C$ .

Try it first, then show the solutionSolution

Let $C = (p, q)$ . $\frac{3 - 2 + p}{3} = 1 \Rightarrow p = 2$ .
$\frac{1 - 4 + q}{3} = - 2 \Rightarrow q - 3 = - 6 \Rightarrow q = - 3$ .

Answer: $C (2, - 3)$

Example 4: Centroid by two routes · 4 marks · Short answerHarder board level

Triangle $A B C$ has $A (5, 7)$ , $B (1, 4)$ and $C (- 3, - 2)$ .
(a) Find the mid-point $D$ of $B C$ .
(b) Find the point dividing $A D$ in the ratio $2 : 1$ .
(c) Show that this is the centroid.

Try it first, then show the solutionSolution

(a) $D = (\frac{1 - 3}{2}, \frac{4 - 2}{2}) = (- 1, 1)$ .
(b) $(\frac{2 ( - 1 ) + 5}{3}, \frac{2 ( 1 ) + 7}{3}) = (1, 3)$ .
(c) $G = (\frac{5 + 1 - 3}{3}, \frac{7 + 4 - 2}{3}) = (1, 3)$ , the same point.

Answer: (a) $(- 1, 1)$ (b) $(1, 3)$ (c) Both give $(1, 3)$ .

Example 5: The mid-point triangle has the same centroid · 4 marks · Short answerHOTS / competency

Triangle $A B C$ has $A (2, 1)$ , $B (8, 3)$ and $C (- 1, 5)$ . $D$ , $E$ , $F$ are the mid-points of $B C$ , $C A$ , $A B$ . Show that triangles $A B C$ and $D E F$ have the same centroid.

Try it first, then show the solutionSolution

Centroid of $A B C$ : $(\frac{2 + 8 - 1}{3}, \frac{1 + 3 + 5}{3}) = (3, 3)$ .
$D (3.5, 4)$ , $E (0.5, 3)$ , $F (5, 2)$ .
Centroid of $D E F$ : $(\frac{3.5 + 0.5 + 5}{3}, \frac{4 + 3 + 2}{3}) = (3, 3)$ , the same.

Answer: Both centroids are $(3, 3)$ .

Example 6: Case study: a triangular plate · 4 marks · Case-basedHOTS / competency

A flat triangular plate has corners $A (6, 9)$ , $B (0, 0)$ and $C (12, 0)$ (in cm).
(a) Find its balance point $G$ .
(b) Find the mid-point $D$ of $B C$ , and show that $G$ divides $A D$ in the ratio $2 : 1$ .
(c) Find the length $A G$ .

Try it first, then show the solutionSolution

(a) $G = (\frac{6 + 0 + 12}{3}, \frac{9 + 0 + 0}{3}) = (6, 3)$ .
(b) $D = (6, 0)$ . $A D$ is vertical: $A G = 9 - 3 = 6$ and $G D = 3 - 0 = 3$ , so $A G : G D = 2 : 1$ .
(c) $A G = 6$ cm.

Answer: (a) $(6, 3)$ (b) $D (6, 0)$ ; $6 : 3 = 2 : 1$ (c) 6 cm

Where students lose marks in Centroid of a Triangle

The slips examiners see most often, and what to write instead.

Mistake: Leaving the centroid unsimplified or not as a point
The answer is a point, simplified.
Wrong working
Centroid $= (\frac{0}{3}, \frac{0}{3})$ , or ' $x = 0$ , $y = 0$ '.
Right working
Centroid $= (0, 0)$ .
Do this: Finish with 'the centroid is $(\dots, \dots)$ '.
Usually costs: Reported in 2024 Q7(iii)(b).
Mistake: Dividing by 2 instead of 3
There are three vertices.
Wrong working
$(\frac{1 + 3 + 5}{2}, \dots)$
Right working
$(\frac{1 + 3 + 5}{3}, \dots) = (3, \dots)$
Do this: Count the points you are averaging.
Usually costs: The accuracy mark.
Mistake: Using the ratio 1 : 2 from the vertex
The centroid is nearer the mid-point: $A G : G D = 2 : 1$ .
Wrong working
$G = (\frac{1 \cdot x _{D} + 2 x _{A}}{3}, \dots)$
Right working
$G = (\frac{2 x _{D} + x _{A}}{3}, \dots)$
Do this: Two parts from the vertex, one part to the side.
Usually costs: Accuracy marks.

Exam tips

2024 Q7(iii)(b) asked for the centroid of a triangle with vertices read from a diagram; the expected answer was $(0, 0)$ , written as a point.
Read vertices from a diagram carefully: a point on the y-axis has $x = 0$ (the 2024 report notes swapped coordinates).
Show the sum for each coordinate before dividing; method marks reward it.
If a question gives a vertex and the mid-point of the opposite side, use the $2 : 1$ section formula directly.

Practice questions on Centroid of a Triangle

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The centroid of the triangle with vertices $(0, 0)$ , $(3, 0)$ and $(0, 6)$ is:
(a) $(1, 2)$
(b) $(3, 6)$
(c) $(1.5, 3)$
(d) $(2, 1)$
Hint
Divide each sum by 3.
Answer
(a) $(1, 2)$
Q2 · 2 marks · Short answerBoard level
Find the centroid of the triangle with vertices $(- 4, - 1)$ , $(2, 5)$ and $(5, - 7)$ .
Hint
Sum of x is 3; sum of y is −3.
Answer
$(1, - 1)$
Q3 · 3 marks · Short answerHarder board level
The centroid of a triangle is $(- 1, 3)$ and two vertices are $(2, 5)$ and $(- 4, 0)$ . Find the third vertex.
Hint
$3 G$ minus the other two.
Answer
$(- 3 - 2 + 4, 9 - 5 - 0) = (- 1, 4)$
Q4 · 3 marks · Short answerHarder board level
The centroid of the triangle with vertices $A (a, - 1)$ , $B (4, b)$ and $C (- 3, 2)$ is $(1, 2)$ . Find $a$ and $b$ .
Hint
$a + 4 - 3 = 3$ and $- 1 + b + 2 = 6$ .
Answer
$a = 2$ , $b = 5$
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The centroid of the triangle with vertices $(1, 1)$ , $(4, 1)$ and $(1, 7)$ is $(2, 3)$ .
Reason (R): The centroid divides each median in the ratio $2 : 1$ from the vertex.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
(A) follows from averaging the vertices.
Answer
(d). Both are true, but (A) is shown by averaging the vertices, not by the median ratio.
Q6 · 3 marks · Short answerHOTS / competency
In triangle $A B C$ , $A$ is $(4, - 6)$ and the mid-point of $B C$ is $D (1, 3)$ . Find the centroid without finding $B$ or $C$ .
Hint
$G$ divides $A D$ in $2 : 1$ .
Answer
$G = (\frac{2 + 4}{3}, \frac{6 - 6}{3}) = (2, 0)$
Q7 · 4 marks · Short answerHarder board level
The centroid of triangle $A B C$ is the origin, $B$ is $(5, - 4)$ and $C$ is $(1, - 2)$ . Find $A$ and the mid-point $D$ of $B C$ , and verify that $A G = 2 G D$ .
Hint
$A = - (B + C)$ .
Answer
$A (- 6, 6)$ , $D (3, - 3)$ . $A G = 72 = 62$ and $G D = 18 = 32$ , so $A G = 2 G D$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Write the centroid formula.Show answer
$(\frac{x _{1} + x _{2} + x _{3}}{3}, \frac{y _{1} + y _{2} + y _{3}}{3})$
2.In what ratio does the centroid divide a median?Show answer
$2 : 1$ from the vertex.
3.Centroid of $(0, 0)$ , $(6, 0)$ , $(0, 3)$ ?Show answer
$(2, 1)$
4.What is a median?Show answer
A segment from a vertex to the mid-point of the opposite side.
5.If $G$ is the origin and two vertices are $(1, 2)$ and $(3, 4)$ , find the third.Show answer
$(- 4, - 6)$

Centroid of a Triangle: frequently asked questions

What is the formula for the centroid of a triangle?

For vertices (x₁, y₁), (x₂, y₂), (x₃, y₃), the centroid is ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3): the average of the three vertices.

Why does the centroid divide a median in 2 : 1?

The three medians meet at one point, two-thirds of the way along each median from its vertex. Using the section formula with ratio 2 : 1 on any median gives the same point as averaging the vertices.

How do I find a missing vertex from the centroid?

Multiply the centroid's coordinates by 3 and subtract the coordinates of the two known vertices.